在上世纪90年代, Larson, Sourour和Kadison在文献[1, 2]中分别独立地提出了局部导子的概念.之后, 学者们开始研究结合代数和非结合代数上的局部导子的结构.设$ \mathscr{H} $是Banach空间, $ \mathscr{B(H)} $是$ \mathscr{H} $上有界线性算子全体构成的代数. Larson和Sourour在文献[1]中证明了$ \mathscr{B(H)} $上的局部导子都是导子; Kadison在文献[2]中证明了Von Neumann代数$ \mathscr{R} $到它的对偶Banach模的任一范数连续的局部导子是导子; 在文献[3]中, 作者证明了$ C^{*} $代数$ \mathfrak{U} $到Banach $ \mathfrak{U} $ -双模上的局部导子是导子; 在文献[4]中, 作者证明了因子von Neumann代数的套子代数的任一范数连续线性局部导子是导子; 赵延霞和王丽在文献[5]中证明了可交换环上上三角矩阵李代数的局部导子是导子; Ayupov和Kudaybergenov在文献[6]中证明了特征为0的代数闭域上有限维半单李代数上的局部导子是导子并且证明了维数大于3的有限维幂零李代数上存在不是导子的局部导子.

2 -局部导子可看作局部导子的一种非线性推广, 它的概念是由Šemrl在文献[7]中提出的.令$ H $是无限维可分离Hilbert空间, $ \mathcal{B}(H) $是$ H $上所有有界线性算子构成的代数. Šemrl在文献[7]中证明了$ \mathcal{B}(H) $上的所有2 -局部导子都是导子; 2011年, Ayupov和Kudaybergenov在文献[8]中将Šemrl的结论推广到任意Hilbert空间上, 即当$ H $是任意Hilbert空间时, $ \mathcal{B}(H) $上的任一2 -局部导子都是导子.此外, Ayupov等人还给出了其他代数上2 -局部导子的结构.例如, 在文献[9]中, 作者们给出了交换正则代数上存在非导子的2 -局部导子的充分必要条件, 同时证明了交换正规代数上矩阵代数的2 -局部导子是导子; 在文献[10]中, 作者们证明了特征为0的代数闭域上有限维半单李代数上的2 -局部导子都是导子, 同时给出了有限维幂零李代数上2 -局部导子不是导子的例子.

令$ R $表示有单位元1的2 -挠自由交换环, $ L_{n}(R) $表示$ R $上所有$ n\times n $反对称矩阵构成的一个李代数.在$ L_{n}(R) $上的李乘定义为$ [A, B] = AB-BA $, 其中$ A, B\in L_{n}(R) $. 2009年, 作者们在文献[11]中证明了$ L_{n}(R)\; (n\geq3) $是完备李代数; 2013年, 文献[12]证明了$ L_{n}(R)(n\geq 5) $上BZ导子的分解式唯一, 进一步得出$ L_{n}(R)\; (n\geq 5) $是完备李代数.由于$ L_{n}(R) $是完备李代数, 所以$ L_{n}(R) $上的导子都是内导子.本文将利用这一结果证明$ L_{n}(R)(n\geq 3) $上的2 -局部导子和局部导子是导子.

定义$ L_{n}(R) $上李乘为$ \left [ A, B \right ] = AB-BA $, $ A, B\in L_{n}(R) $.设$ E_{ij} $为$ (i, j) $位置为1, 其余位置为0的$ n $阶方阵.令$ A_{ij} = E_{ij}-E_{ji} $, 则$ A_{ij}\in L_{n}(R) $, $ A_{ij} = -A_{ji} $, $ A_{ii} = 0 $.根据$ L_{n}(R) $上李乘的定义可知

$ \left [ A_{ij}, A_{sk}\right] = \delta_{ik}A_{js}+\delta_{js}A_{ik}-\delta_{is}A_{jk}-\delta_{jk}A_{is}, 1\leq i, j, k, s\leq n. $

本文选择$ \left \{ A_{kl}|{1\leqslant k<l\leqslant n} \right \} $作为$ L_{n}\left ( R \right ) $的一组基底.设$ X = \sum\limits_{1\leq k<l\leq n}x_{kl}A_{kl}\in L_{n}(R) $, $ x_{kl}\in R $, 那么

$ [X, A_{ij}] = \sum\limits_{1\leq k<l\leq n}x_{kl}[A_{kl}, A_{ij}] = \sum\limits_{k = 1}^{i-1}x_{ki}A_{kj}-\sum\limits_{l = i+1, l\neq j}^{n}x_{il}A_{lj}- \sum\limits_{k = 1, k\neq i}^{j-1}x_{kj}A_{ki}+\sum\limits_{l = j+1}^{n}x_{jl}A_{li}. $

定义2.1^[13]  令$ R $是有1的交换环, $ L $是$ R $上的李代数.若$ R $ -线性映射$ \phi :L\rightarrow L $满足

$ \phi \left [ x, y \right ] = \left [ \phi (x), y \right ]+\left [ x, \phi(y) \right ], x, y \in L, $

则称$ \phi $为导子.特别的, 任意$ z\in L $, $ \mathrm{ad}z:L\rightarrow L $, $ \mathrm{ad}z(x) = [z, x], x\in L $也是一个导子, 称这种形式的导子为内导子.

定义2.2^[14]  令$ R $是有1的交换环, $ L $是$ R $上的李代数.若满足

(ⅰ) $ Z(L) = \{x\in L|[x, y] = 0, y\in L\} = 0 $,

(ⅱ) $ L $的导子都是内导子,

则称$ L $是完备李代数.

命题2.3 (见文献[11, Theorem 3.2])令$ R $是有1的2 -挠自由交换环, 那么$ L_{n}(R)\; (n\geq3) $是完备李代数.

定义2.4^[6]  令$ R $是有1的交换环, $ L $是$ R $上的李代数.若$ R $ -线性映射$ \Delta :L\rightarrow L $满足:对于任意$ x \in L $都存在一个导子$ D_{x} $$ ( $与$ x $有关$ ) $使得$ \Delta(x) = D_{x}(x) $, 则称$ \Delta $为$ L $的局部导子.

定义2.5^[10]  令$ R $是有1的交换环, $ L $是$ R $上的李代数.若映射$ \mathrm{T}:L\rightarrow L $满足:对于任意$ x, y \in L $都存在一个导子$ D_{x, y} $$ ( $与$ x, y $有关$ ) $使得$ \mathrm{T}(x) = D_{x, y}(x) $, $ \mathrm{T}(y) = D_{x, y}(y) $, 则称$ \mathrm{T} $为$ L $的2 -局部导子.

显然, 导子是局部导子, 也是2 -局部导子.反之, 不一定成立.

下文设$ \mathrm{T} $和$ \Delta $分别是$ L_{n}(R) $上的2 -局部导子和局部导子.

由命题2.3直接可得$ L_{n}(R)\; (n\geq 3) $上的导子都是内导子, 因此对于任意$ A, B\in L_{n}(R) $, 存在$ M_{A, B}\in L_{n}(R) $, 满足$ \mathrm{T}(x) = [M_{A, B}, A] $, $ \mathrm{T}(y) = [M_{A, B}, B] $.对于任意$ A\in L_{n}(R) $, 存在$ M_{A}\in L_{n}(R) $, 满足$ \Delta(x) = [M_{A}, A] $.

根据局部导子定义可得, 存在$ A\in L_{n}(R)\; (n\geq3) $使得$ \Delta(A_{12}) = [A, A_{12}] $.令$ \Delta^{'} = \Delta-\mathrm{ad}A $, 则$ \Delta^{'}(A_{12}) = 0 $.显然, 如果$ \Delta^{'} $是导子, 那么$ \Delta $是导子.下文中用$ \Delta $表示$ \Delta^{'} $, 并将证明$ \Delta $是一个导子.

引理3.1  设$ \Delta $是$ L_{3}(R) $上的局部导子.若$ \Delta(A_{12}) = 0 $, 则$ \Delta $是导子.

证  显然, $ A_{12}, A_{13}, A_{23} $可作为$ L_{3}(R) $的一组基.根据局部导子的定义, 存在$ x, y\in L_{3}(R) $, 使得$ \Delta(A_{13}) = [x, A_{13}] $, $ \Delta(A_{23}) = [y, A_{23}] $.设

$ x = x_{1}A_{12}+x_{2}A_{13}+x_{3}A_{23}, \; \; y = y_{1}A_{12}+y_{2}A_{13}+y_{3}A_{23}, $

那么$ \Delta(A_{13}) = -x_{1}A_{23}+x_{3}A_{12} $, $ \Delta(A_{23}) = y_{1}A_{13}-y_{2}A_{12} $.

对$ A_{12}+A_{13} $, 存在$ m\in L_{3}(R) $, $ m = m_{1}A_{12}+m_{2}A_{13}+m_{3}A_{23} $, 使得$ \Delta(A_{12}+A_{13}) = [m, A_{12}+A_{13}] $.因为$ \Delta $是线性映射, 所以$ \Delta(A_{12}+A_{13}) = \Delta(A_{12})+\Delta(A_{13}) $, 即$ [m, A_{12}+A_{13}] = 0+[x, A_{13}] = [x, A_{13}] $.将$ m $和$ x $的基底表示代入等式, 整理得$ x_{3} = m_{3} = 0 $.同理, 对$ A_{12}+A_{23} $和$ A_{13}+A_{23} $分别进行如上操作可得$ y_{2} = 0 $和$ x_{1} = y_{1} $.所以

$ \Delta(A_{13}) = -x_{1}A_{23} = [x_{1}A_{12}, A_{13}], \; \; \Delta(A_{23}) = y_{1}A_{13} = x_{1}A_{13} = [x_{1}A_{12}, A_{23}]. $

令$ \Delta_{1} = \Delta-\mathrm{ad}(x_{1}A_{12}) $, 那么$ \Delta_{1}(A_{12}) = \Delta_{1}(A_{13}) = \Delta_{1}(A_{23}) = 0 $, 所以$ \Delta_{1} = 0 $, 即$ \Delta = \mathrm{ad}(x_{1}A_{12}) $.因此$ L_{3}(R) $上的局部导子$ \Delta $是导子.

在$ L_{n}(R)(n\geq4) $中, 当$ 3\leq j\leq n $时, 由局部导子定义可得, 存在

$ a^{(j)} = \sum\limits_{1\leq k<l\leq n}a_{kl}^{(j)}A_{kl}\in L_{n}(R), \; \; b^{(j)} = \sum\limits_{1\leq k<l\leq n}b_{kl}^{(j)}A_{kl}\in L_{n}(R). $

使得

$ \begin{eqnarray} &&\Delta(A_{1j}) = [a^{(j)}, A_{1j}] = \sum\limits_{k = 2}^{n}a_{1k}^{(j)}A_{jk}+\sum\limits_{k = 2}^{j-1}a_{kj}^{(j)}A_{1k}- \sum\limits_{k = j+1}^{n}a_{jk}^{(j)}A_{1k}, \end{eqnarray} $

(3.1)

$ \begin{eqnarray} &&\Delta(A_{2j}) = [b^{(j)}, A_{2j}] = b_{12}^{(j)}A_{1j}+\sum\limits_{k = 3}^{n}b_{2k}^{(j)}A_{jk}+\sum\limits_{k = 1}^{j-1}b_{kj}^{(j)}A_{2k}- \sum\limits_{k = j+1}^{n}b_{jk}^{(j)}A_{2k}. \end{eqnarray} $

(3.2)

引理3.2  设$ \Delta $是$ L_{n}(R)(n\geq4) $上的局部导子.若$ \Delta(A_{12}) = 0 $, 那么存在$ m\in L_{n}(R) $, 令$ \Delta_{1} = \Delta-\mathrm{ad}m $, 使得

$ \begin{eqnarray*} &&\Delta _{1}(A_{1j}) = \sum\limits_{k = 3}^{j-1}a_{kj}^{(j)}A_{1k}-\sum\limits_{k = j+1}^{n}a_{jk}^{(j)}A_{1k}, j = 3, 4, \cdots, n;\\ &&\Delta _{1}(A_{2j}) = \sum\limits_{k = 3}^{j-1}a_{kj}^{(j)}A_{2k}-\sum\limits_{k = j+1}^{n}a_{jk}^{(j)}A_{2k}, j = 3, 4, \cdots, n;\\ &&\Delta _{1}(A_{ij}) = \sum\limits_{k = 3}^{i-1}a_{ki}^{(i)}A_{kj}-\sum\limits_{k = i+1, k\neq j}^{n}a_{ik}^{(i)}A_{kj}-\sum\limits_{k = 3, k\neq i}^{j-1}a_{kj}^{(j)}A_{ki}+\sum\limits_{k = j+1}^{n}a_{jk}^{(j)}A_{ki}, 3\leq i<j\leq n, \end{eqnarray*} $

且当$ 3\leq i<j\leq n $时, $ a_{ij}^{(j)} = a_{ij}^{(i)} $.

证  在引理3.2的证明中, 始终假设$ 3\leq i, j\leq n $.

根据局部导子定义, 存在$ x = \sum\limits_{1\leq k<l\leq n}x_{kl}A_{kl}\in L_{n}(R) $, $ y = \sum\limits_{1\leq k<l\leq n}y_{kl}A_{kl}\in L_{n}(R) $, 使得$ \Delta(A_{12}+A_{1j}) = [x, A_{12}+A_{1j}] $, $ \Delta(A_{12}+A_{2j}) = [y, A_{12}+A_{2j}] $.因为$ \Delta $是线性映射, 所以

$ \Delta(A_{12}+A_{1j}) = \Delta(A_{12})+\Delta(A_{1j}) = \Delta(A_{1j}), \Delta(A_{12}+A_{2j}) = \Delta(A_{12})+\Delta(A_{2j}) = \Delta(A_{2j}). $

即$ [x, A_{12}+A_{1j}] = [a^{(j)}, A_{1j}] $, $ [y, A_{12}+A_{2j}] = [b^{(j)}, A_{2j}], $

$ \begin{align*} [x, A_{12}+A_{1j}]& = \sum\limits_{1\leq k<l\leq n}x_{kl}[A_{kl}, A_{12}]+\sum\limits_{1\leq k<l\leq n}x_{kl}[A_{kl}, A_{1j}] \\ & = \sum\limits_{k = 3}^{n}x_{1k}A_{2k}-\sum\limits_{k = 3}^{n}x_{2k}A_{1k}+\sum\limits_{k = 2}^{n}x_{1k}A_{jk}+\sum\limits_{k = 2}^{j-1}x_{kj}A_{1k}-\sum\limits_{k = j+1}^{n}x_{jk}A_{1k} \\ & = \sum\limits_{k = 2}^{n}a_{1k}^{(j)}A_{jk}+\sum\limits_{k = 2}^{j-1}a_{kj}^{(j)}A_{1k}- \sum\limits_{k = j+1}^{n}a_{jk}^{(j)}A_{1k} \\ & = [a^{(j)}, A_{1j}], \end{align*} $

所以$ a_{1k}^{(j)} = x_{1k}, k = 3, \cdots, n, k\neq j $; $ x_{1k} = 0, k = 3, \cdots, n, k\neq j $, 即$ a_{1k}^{(j)} = 0, k = 3, \cdots, n, k\neq j $.同理$ b_{2k}^{(j)} = y_{2k}, k = 3, \cdots, n, k\neq j $; $ y_{2k} = 0, k = 3, \cdots, n, k\neq j $, 即$ b_{2k}^{(j)} = 0, k = 3, \cdots, n, k\neq j $.以上结果代入式(3.1), (3.2)式得

$ \begin{eqnarray} &&\Delta (A_{1j}) = \sum\limits_{k = 3}^{j-1}a_{kj}^{(j)}A_{1k}-\sum\limits_{k = j+1}^{n}a_{jk}^{(j)}A_{1k}+a_{2j}^{(j)}A_{12}-a_{12}^{(j)}A_{2j}, j = 3, 4, \cdots, n, \end{eqnarray} $

(3.3)

$ \begin{eqnarray} &&\Delta (A_{2j}) = \sum\limits_{k = 3 }^{j-1}b_{kj}^{(j)}A_{2k}-\sum\limits_{k = j+1 }^{n}b_{jk}^{(j)}A_{2k}-b_{1j}^{(j)}A_{12}+b_{12}^{(j)}A_{1j}, j = 3, 4, \cdots, n. \end{eqnarray} $

(3.4)

存在$ d = \sum\limits_{1\leq k<l\leq n}d_{kl}A_{kl}\in L_{n}(R) $, 使得$ \Delta(A_{1j}+A_{1i}) = [d, A_{1j}+A_{1i}] $, 不妨设$ i<j $.又因为$ \Delta(A_{1j}+A_{1i}) = \Delta(A_{1j})+\Delta(A_{1i}) $, 即$ [d-a^{(j)}, A_{1j}] = [a^{(i)}-d, A_{1i}] $, 所以

$ \begin{equation} a_{12}^{(i)} = a_{12}^{(j)}, a_{ij}^{(i)} = a_{ij}^{(j)}, 3\leq i<j\leq n. \end{equation} $

(3.5)

存在$ h = \sum\limits_{1\leq k<l\leq n}h_{kl}A_{kl}\in L_{n}(R) $, 满足$ \Delta(A_{1j}+A_{2j}) = [h, A_{1j}+A_{2j}] $.同理可得$ [h-a^{(j)}, A_{1j}] = [b^{(j)}-d, A_{2j}], $所以

$ \begin{eqnarray} &&a_{12}^{(j)} = b_{12}^{(j)}, \end{eqnarray} $

(3.6)

$ \begin{eqnarray} &&a_{kj}^{(j)} = b_{kj}^{(j)}, k = 3, \cdots, j-1, \end{eqnarray} $

(3.7)

$ \begin{eqnarray} &&a_{jk}^{(j)} = b_{jk}^{(j)}, k = j+1, \cdots, n. \end{eqnarray} $

(3.8)

存在$ z = \sum\limits_{1\leq k<l\leq n}z_{kl}A_{kl}\in L_{n}(R) $, 满足$ \Delta(A_{1j}+A_{2i}) = [z, A_{1j}+A_{2i}] $, 假设$ i\neq j $.同理可得$ [z, A_{1j}+A_{2i}] = [a^{j}, A_{1j}]+[b^{(i)}, A_{2i}] $.在上式展开式中有$ (z_{2j}-z_{1i})A_{12} = (a_{2j}^{(j)}-b_{1i}^{(i)})A_{12} $和$ (z_{2j}-z_{1i})A_{ij} = 0 $.所以$ z_{2j}-z_{1i} = a_{2j}^{(j)}-b_{1i}^{(i)} $, $ z_{2j}-z_{1i} = 0 $.因此

$ \begin{eqnarray} a_{2j}^{(j)} = b_{1i}^{(i)}, 3\leq i\neq j\leq n. \end{eqnarray} $

(3.9)

由式(3.5)第一个式子和式(3.6)可得

$ \begin{equation} a_{12}^{(j)} = b_{12}^{(i)}, i, j = 3, \cdots, n. \end{equation} $

(3.10)

因此不妨设$ a = a_{12}^{(j)} $.

令$ m = aA_{12} $, $ \Delta_{1} = \Delta-\mathrm{ad}m $.将(3.7)和(3.8)式分别代入(3.3)和(3.4)式, 得

$ \begin{eqnarray} &&\Delta _{1}(A_{1j}) = \sum\limits_{k = 3}^{j-1}a_{kj}^{(j)}A_{1k}-\sum\limits_{k = j+1}^{n}a_{jk}^{(j)}A_{1k}+a_{2j}^{(j)}A_{12}, j = 3, 4, \cdots, n, \end{eqnarray} $

(3.11)

$ \begin{eqnarray} &&\Delta _{1}(A_{2j}) = \sum\limits_{k = 3}^{j-1}a_{kj}^{(j)}A_{2k}-\sum\limits_{k = j+1}^{n}a_{jk}^{(j)}A_{2k}-b_{1j}^{(j)}A_{12}, j = 3, 4, \cdots, n. \end{eqnarray} $

(3.12)

此时$ \Delta _{1}(A_{12}) = 0 $, 且$ \Delta_{1} $仍然是一个局部导子.

存在$ c = \sum\limits_{1\leq k<l\leq n}c_{kl}A_{kl}\in L_{n}(R) $, 使得

$ \begin{equation} \begin{aligned} \Delta_{1}(A_{ij})& = [c, A_{ij}] \\ & = \sum\limits_{k = 1}^{i-1}c_{ki}A_{kj}-\sum\limits_{k = i+1}^{n}c_{ik}A_{kj} -\sum\limits_{k = 1}^{j-1}c_{kj}A_{ki}+\sum\limits_{k = j+1}^{n}c_{jk}A_{ki}, 3\leq i<j\leq n. \end{aligned} \end{equation} $

(3.13)

存在$ t = \sum\limits_{1\leq k<l\leq n}t_{kl}A_{kl}\in L_{n}(R) $, 满足$ \Delta_{1}(A_{12}+A_{ij}) = [t, A_{12}+A_{ij}] $.因为$ \Delta_{1} $是线性映射, 所以$ \Delta_{1}(A_{12}+A_{ij}) = \Delta_{1}(A_{12})+\Delta(A_{ij}) = \Delta(A_{ij}) $, 即$ [t, A_{12}+A_{ij}] = [c, A_{ij}] $.在展开式中有$ c_{1i}A_{1j} = (t_{1i}-t_{2j})A_{1j} $, $ c_{2i}A_{2j} = (t_{1j}+t_{2i})A_{2j} $, $ c_{1j}A_{1i} = (t_{2i}+t_{1j})A_{1i} $, $ -c_{2j}A_{2i} = (t_{1i}-t_{2j})A_{2i} $, 所以$ c_{1i} = t_{1i}-t_{2j} $, $ c_{2i} = t_{1j}+t_{2i} $, $ c_{1j} = t_{2i}+t_{1j} $, $ -c_{2j} = t_{1i}-t_{2j} $.因此

$ \begin{equation} c_{1i} = -c_{2j}, c_{2i} = c_{1j}. \end{equation} $

(3.14)

存在$ u = \sum\limits_{1\leq k<l\leq n}u_{kl}A_{kl}\in L_{n}(R) $, 满足$ \Delta_{1}(A_{1j}+A_{ij}) = [u, A_{1j}+A_{ij}] $.因为$ \Delta_{1} $是线性映射, 所以$ \Delta_{1}(A_{1j}+A_{ij}) = \Delta_{1}(A_{1j})+\Delta_{1}(A_{ij}) $, 即$ [u, A_{1j}+A_{ij}] = [c, A_{ij}]+[a^{(j)}-m, A_{1j}] $.式子展开后可得

$ \begin{eqnarray} &&a_{2j}^{(j)} = u_{2j} = c_{2j}; \end{eqnarray} $

(3.15)

$ \begin{eqnarray} &&-c_{1i} = -u_{1i} = 0; \end{eqnarray} $

(3.16)

$ \begin{eqnarray} &&a_{kj}^{(j)} = u_{kj} = c_{kj}, k = 3, \cdots, j-1, k\neq i; \end{eqnarray} $

(3.17)

$ \begin{eqnarray} &&a_{jk}^{(j)} = u_{jk} = c_{jk}, k = j+1, \cdots, n. \end{eqnarray} $

(3.18)

由(3.9), (3.15)和(3.16)式可知, $ a_{2j}^{(j)} = b_{1j}^{(j)} = 0, 3\leq j\leq n $.

同理, 存在$ v = \sum\limits_{1\leq k<l\leq n}v_{kl}A_{kl}\in L_{n}(R) $, 满足$ \Delta_{1}(A_{1i}+A_{ij}) = [v, A_{1i}+A_{ij}] $.因为$ \Delta_{1} $是线性映射, 所以$ \Delta_{1}(A_{1i}+A_{ij}) = \Delta_{1}(A_{1i})+\Delta_{1}(A_{ij}) $, 即

$ [v, A_{1i}+A_{ij}] = [c, A_{ij}]+[a^{(i)}-m, A_{1i}]. $

式子展开后可得

$ \begin{eqnarray} &&c_{1j} = v_{1j} = 0; \end{eqnarray} $

(3.19)

$ \begin{eqnarray} &&c_{ki} = v_{ki} = a_{ki}^{(i)}, k = 3, \cdots, i-1, k\neq i; \end{eqnarray} $

(3.20)

$ \begin{eqnarray} &&c_{ik} = v_{ik} = a_{ik}^{(i)}, k = i+1, \cdots, n. \end{eqnarray} $

(3.21)

将(3.14)–(3.21)式代入(3.11)–(3.13)式, 可得

$ \begin{eqnarray*} &&\Delta _{1}(A_{1j}) = \sum\limits_{k = 3}^{j-1}a_{kj}^{(j)}A_{1k}-\sum\limits_{k = j+1}^{n}a_{jk}^{(j)}A_{1k}, j = 3, 4, \cdots, n;\\ &&\Delta _{1}(A_{2j}) = \sum\limits_{k = 3}^{j-1}a_{kj}^{(j)}A_{2k}-\sum\limits_{k = j+1}^{n}a_{jk}^{(j)}A_{2k}, j = 3, 4, \cdots, n;\\ &&\Delta _{1}(A_{ij}) = \sum\limits_{k = 3}^{i-1}a_{ki}^{(i)}A_{kj}-\sum\limits_{k = i+1, k\neq j}^{n}a_{ik}^{(i)}A_{kj}-\sum\limits_{k = 3, k\neq i}^{j-1}a_{kj}^{(j)}A_{ki}+\sum\limits_{k = j+1}^{n}a_{jk}^{(j)}A_{ki}, 3\leq i<j\leq n. \end{eqnarray*} $

引理证毕.

定理3.3  $ L_{n}(R)(n\geq 3) $上的局部导子$ \Delta $是导子.

证  由引理3.1可知, $ L_{3}(R) $上的局部导子都是导子, 所以只需证明$ n\geq4 $时结论成立.

因为$ \Delta_{1} $是线性映射, 所以如果对基底中任意两个元素$ A_{ij} $, $ A_{kl} $, $ 1\leq i, j, k, l\leq n $都满足以下等式

$ [\Delta _{1}(A_{ij}), A_{kl}]+[A_{ij}, \Delta _{1}(A_{kl})] = \Delta _{1}([A_{ij}, A_{kl}]). $

则对任意$ x\in L_{n}(R) $, $ y\in L_{n}(R) $, 都有$ [\Delta _{1}(x), y]+[x, \Delta _{1}(y)] = \Delta _{1}([x, y]) $成立, 即$ \Delta _{1} $是导子.现将基底分为四部分: $ A_{12} $, $ \left\{A_{1p}\right\}_{3\leq p\leq n} $, $ \left\{A_{2p}\right\}_{3\leq p\leq n} $, $ \left\{A_{pq}\right\}_{3\leq p<q\leq n} $.因此只需验证以下10个等式成立即可,

$ \begin{eqnarray*} &&[\Delta _{1}(A_{12}), A_{ij}]+[A_{12}, \Delta _{1}(A_{ij})] = 0;\\ && [\Delta _{1}(A_{12}), A_{2j}]+[A_{12}, \Delta _{1}(A_{2j})] = \Delta _{1}(A_{1j});\\ && [\Delta _{1}(A_{12}), A_{1j}]+[A_{12}, \Delta _{1}(A_{1j})] = -\Delta _{1}(A_{2j});\\ &&[\Delta _{1}(A_{1i}), A_{2j}]+[A_{1i}, \Delta _{1}(A_{2j})] = 0;\\ &&[\Delta _{1}(A_{1j}), A_{2j}]+[A_{1j}, \Delta _{1}(A_{2j})] = 0;\\ &&[\Delta _{1}(A_{ij}), A_{kl}]+[A_{ij}, \Delta _{1}(A_{kl})] = 0;\\ &&[\Delta _{1}(A_{ij}), A_{il}]+[A_{ij}, \Delta _{1}(A_{il})] = -\Delta _{1}(A_{jl});\\ &&[\Delta _{1}(A_{sj}), A_{kl}]+[A_{sj}, \Delta _{1}(A_{kl})] = 0, s = 1, 2;\\ &&[\Delta _{1}(A_{si}), A_{ij}]+[A_{si}, \Delta _{1}(A_{ij})] = \Delta _{1}(A_{sj}), s = 1, 2;\\ &&[\Delta _{1}(A_{si}), A_{sj}]+[A_{si}, \Delta _{1}(A_{sj})] = -\Delta _{1}(A_{ij}), s = 1, 2, \end{eqnarray*} $

其中$ 3\leq i, j, k, l\leq n $, 且$ i, j, k, l $互不相等.

以下验证第7个等式成立, 不妨假设$ i<j<l $,

$ \begin{align*} [\Delta _{1}(A_{ij}), A_{il}] & = -a_{il}^{(i)}[A_{lj}, A_{il}] -\sum\limits_{k = 3, k\neq i}^{j-1}a_{kj}^{(j)}[A_{ij}, A_{il}] +\sum\limits_{k = j+1}^{n}a_{jk}^{(j)}[A_{ki}, A_{il}]\\ & = a_{il}^{(l)}A_{ij} -\sum\limits_{k = 3, k\neq i}^{j-1}a_{kj}^{(j)}A_{kl} +\sum\limits_{k = j+1, k\neq l}^{n}a_{jk}^{(j)}A_{kl}, \\ [A_{ij}, \Delta _{1}(A_{il})] & = -a_{ij}^{(i)}[A_{ij}, A_{jl}] -\sum\limits_{k = 3, k\neq i}^{l-1}a_{kl}^{(l)}[A_{ij}, A_{ki}] +\sum\limits_{k = l+1}^{n}a_{lk}^{(l)}[A_{ij}, A_{ki}]\\ & = -a_{ij}^{(j)}A_{il} -\sum\limits_{k = 3, k\neq i, j}^{l-1}a_{kl}^{(l)}A_{jk} +\sum\limits_{k = l+1}^{n}a_{lk}^{(l)}A_{jk}, \end{align*} $

所以

$ \begin{align*} &[\Delta _{1}(A_{ij}), A_{il}]+[A_{ij}, \Delta_{1}(A_{il})]\\ = &-\sum\limits_{k = 3}^{j-1}a_{kj}^{(j)}A_{kl}+\sum\limits_{k = j+1, k\neq l}^{n}a_{jk}^{(j)}A_{kl}+\sum\limits_{k = 3, k\neq j}^{l-1}a_{kl}^{(l)}A_{kj}-\sum\limits_{k = l+1}^{n}a_{lk}^{(l)}A_{kj}, \\ = &-\Delta _{1}(A_{jl}). \end{align*} $

由以上可得等式$ [\Delta _{1}(A_{ij}), A_{il}]+[A_{ij}, \Delta _{1}(A_{il})] = -\Delta _{1}(A_{jl}) $成立, 其余等式类似可得.因此$ \Delta_{1} $是导子, 根据命题2.3可知存在$ X\in L_{n}(R) $, 使得$ \Delta_{1} = \mathrm{ad}X $.所以

$ \Delta = \Delta _{1}+\mathrm{ad}m = \mathrm{ad}(X+m), $

即$ \Delta $是一个导子.定理得证.

引理4.1  $ L_{n}(R)\; (n\geq 3) $上的2 -局部导子是线性映射.

证  用$ \mathrm{tr}(X) $表示矩阵$ X $的迹, 即$ X $的对角元素之和.令$ f(A, B) = \mathrm{tr}(AB), A, B\in L_{n}(R) $, 则

$ \begin{eqnarray*} &&f(A, B) = f(B, A), \\ &&f(A_{1}+A_{2}, B_{1}+B_{2}) = f(A_{1}, B_{1})+f(A_{1}, B_{2})+f(A_{2}, B_{1})+f(A_{2}, B_{2}), \\ &&f(rA, tB) = rtf(A, B), \\ &&f([A, B], C) = f(A, [B, C]), \end{eqnarray*} $

其中$ A, B, A_{1}, A_{2}, B_{1}, B_{2}\in L_{n}(R) $, $ r, t\in R $.

若$ x = \sum\limits_{1\leq k<l\leq n}x_{kl}A_{kl}\in L_{n}(R) $满足$ f(A, x) = 0 $, $ \forall A\in L_{n}(R) $, 那么$ f(x, A_{ij}) = -2x_{ij} = 0 $, $ 1\leq i<j \leq n $, 所以$ x_{ij} = 0 $, 即$ x = 0 $.

由以上可得$ f $是$ L_{n}(R) $上非退化的对称双线性型.

设$ \mathrm{T} $是$ L_{n}(R)(n\geq 3) $上的2 -局部导子, 那么对于任意$ A, B, C\in L_{n}(R) $, 有

$ \begin{align*} f(\mathrm{T}(A+B), C) & = f([M_{A+B, C}, A+B], C) = -f(A+B, [M_{A+B, C}, C]) \\ & = -f(A+B, \mathrm{T}(C)) = -f(A, \mathrm{T}(C))-f(B, \mathrm{T}(C))\\ & = -f(A, [M_{A, C}, C])-f(B, [M_{B, C}, C]) \\ & = f([M_{A, C}, A], C)+f([M_{B, C}, B], C)\\ & = f(\mathrm{T}(A), C)+f(\mathrm{T}(B), C) \\ & = f(\mathrm{T}(A)+\mathrm{T}(B), C), \end{align*} $

所以$ \mathrm{T}(A+B) = \mathrm{T}(A)+\mathrm{T}(B) $.

对于任意$ A\in L_{n}(R), r\in R $, 存在$ M_{A, rA} $使得$ \mathrm{T}(A) = [M_{A, rA}, A] $, $ \mathrm{T}(rA) = [M_{A, rA}, rA] $，显然$ \mathrm{T}(rA) = r\mathrm{T}(A) $.因此$ \mathrm{T} $是线性映射.引理得证.

定理4.2  $ L_{n}(R)\; (n\geq 3) $上的2 -局部导子是导子.

证  由引理4.1知$ T $具有线性性, 因此只需证明$ [\mathrm{T}(A_{ij}), A_{kl}]+[A_{ij} $, $ \mathrm{T}(A_{kl})] = \mathrm{T}([A_{ij}, A_{kl}]) $成立即可.

分两种情况证明.

(ⅰ)当$ [A_{ij}, A_{kl}] = 0 $时, 根据2 -局部导子定义, 存在$ m\in L_{n}(R) $, 使得$ \mathrm{T}(A_{ij}) = [m, A_{ij}] $, $ \mathrm{T}(A_{kl}) = [m, A_{kl}] $, 那么

$ [\mathrm{T}(A_{ij}), A_{kl}]+[A_{ij}, \mathrm{T}(A_{kl})] = [m, [A_{ij}, A_{kl}]] = 0 = \mathrm{T}([A_{ij}, A_{kl}]). $

等式成立.

(ⅱ)当$ [A_{ij}, A_{kl}]\neq 0 $时, 不妨写成$ [A_{ij}, A_{is}] = -A_{js} $, 其中$ i<j<s $.根据2 -局部导子的定义, 存在

$ x = \sum\limits_{1\leq k<l\leq n}x_{kl}A_{kl}, \; \; y = \sum\limits_{1\leq k<l\leq n}y_{kl}A_{kl}, \; \; m = \sum\limits_{1\leq k<l\leq n}m_{kl}A_{kl}\in L_{n}(R), $

满足$ \mathrm{T}(A_{ij}) = [x, A_{ij}] = [m, A_{ij}] $, $ \mathrm{T}(A_{is}) = [y, A_{is}] = [m, A_{is}] $, $ \mathrm{T}(A_{js}) = [x, A_{js}] = [y, A_{js}] $.因此可得

$ \begin{align*} \begin{split} 0 = [x-m, A_{ij}] = &\sum\limits_{k = 1}^{i-1}(x_{ki}-m_{ki})A_{kj}-\sum\limits_{k = i+1, k\neq j}^{n}(x_{ik}-m_{ik})A_{kj}\\ &-\sum\limits_{k = 1, k\neq i}^{j-1}(x_{kj}-m_{kj})A_{ki}+\sum\limits_{k = j+1}^{n}(x_{jk}-m_{jk})A_{ki}, \\ 0 = [y-m, A_{is}] = &\sum\limits_{k = 1}^{i-1}(y_{ki}-m_{ki})A_{ks}-\sum\limits_{k = i+1, k\neq s}^{n}(y_{ik}-m_{ik})A_{ks}\\ &-\sum\limits_{k = 1, k\neq i}^{s-1}(y_{ks}-m_{ks})A_{ki}+\sum\limits_{k = s+1}^{n}(y_{sk}-m_{sk})A_{ki}, \\ \end{split} \end{align*} $

$ \begin{align*} \begin{split} 0 = [x-y, A_{js}] = &\sum\limits_{k = 1}^{j-1}(x_{kj}-y_{kj})A_{ks}-\sum\limits_{k = j+1, k\neq s}^{n}(x_{jk}-y_{jk})A_{ks}\\ &-\sum\limits_{k = 1, k\neq j}^{s-1}(x_{ks}-y_{ks})A_{kj}+\sum\limits_{k = s+1}^{n}(x_{sk}-y_{sk})A_{kj}. \end{split} \end{align*} $

将上式各项展开可得

$ \begin{eqnarray*} &&x_{ki}-m_{ki} = 0, k = 1, \cdots, i-1; x_{ik}-m_{ik} = 0, k = i+1, \cdots, n, k\neq j;\\ &&x_{kj}-m_{kj} = 0, k = 1, \cdots, j-1, k\neq i; x_{jk}-m_{jk} = 0, k = j+1, \cdots, n;\\ &&y_{ki}-m_{ki} = 0, k = 1, \cdots, i-1; y_{ik}-m_{ik} = 0, k = i+1, \cdots, n, k\neq s;\\ &&y_{ks}-m_{ks} = 0, k = 1, \cdots, s-1, k\neq i; y_{sk}-m_{sk} = 0, k = s+1, \cdots, n;\\ &&x_{kj}-y_{kj} = 0, k = 1, \cdots, j-1; x_{jk}-y_{jk} = 0, k = j+1, \cdots, n, k\neq s.\\ &&x_{ks}-y_{ks} = 0, k = 1, \cdots, s-1, k\neq j; x_{sk}-y_{sk} = 0, k = s+1, \cdots, n; \end{eqnarray*} $

所以

$ \begin{eqnarray*} &&x_{kj} = m_{kj}, k = 1, \cdots, j-1; x_{jk} = m_{jk}, k = j+1, \cdots, n, k\neq s;\\ &&x_{ks} = m_{ks}, k = 1, \cdots, s-1, k\neq j; x_{sk} = m_{sk}, k = s+1, \cdots, n. \end{eqnarray*} $

即$ [x-m, A_{js}] = 0. $所以$ \mathrm{T}(A_{js}) = [x, A_{js}] = [m, A_{js}]. $因此

$ [\mathrm{T}(A_{ij}), A_{is}]+[A_{ij}, \mathrm{T}(A_{is})] = [m, [A_{ij}, A_{is}]] = -\mathrm{T}(A_{js}). $

定理得证.